0:06

For a function f(x), differentiable sufficiently many times,

we say that a differentiable polynomial P(D) annihilates

f(x) if action of P(D) on f is equal to 0, okay?

In this case, we call the differential polynomial P(D) an annihilator of f(x),

or sometimes a differential polynomial annihilator of f(x), okay?

For example, for any positive integer n,

D to the n annihilates x to the k, for all k from 0 to n-1.

But that same operator D to the n,

it does not annihilates x to the k, for

k is greater than or equal to n, okay?

So in fact, D to the n annihilates any polynomial of degree

strictly less than n, okay?

1:15

For example, I'd like to find annihilators of the following functions, okay?

First, 3 times e to the -x- 2 times x times e to the 2x, okay?

How to find this annihilator, okay?

Let's look at the term by term then, okay?

First, D + 1 e to the -x equals to 0, right?

I hope you can see it easily.

Again D minus 2 e to the 2x is equals to 0 ,right, okay?

I think you know that D minus 2 annihilates e to the 2x, right, okay?

Okay, what happen if we multiply e to the 2x by x then?

Okay, it will be annihilated by (D-2) square, right?

Okay, can you see it, right?

So then combining this two,

(D+1)e to the -x = 0, and

(D-2 )square x e to the 2x =0, okay?

So that (D+1) times (D-2) square, okay, it annihilates the whole expression.

3 times e to the negative x- 2 time x e to the 2x, right?

So this is an annihilator, okay?

This differential polynomial of order 3,

this is an annihilator of the given expression, okay?

For the second example, -2e to the -x sine 2x, right?

You can recognize e to the -x sine

of 2x as an imaginary part of exponential

-1 plus 2i of x, right, okay?

And you also know that, okay, D-(-1 +2i)

annihilate exponential (-1+2i)/x, right?

Okay, similarly, okay, to find the annihilator of this e to the -x sine 2x,

okay, we're seeking quadratic polynomial in D, okay?

We are seeking a quadratic polynomial in D,

having -1 +- 2i, as its two roots,

okay, which is trivially D-(-1+2i)

times D-(-1-2i), okay?

Expanding this product out,

you are going to get D square+2D+5, okay?

Then I claim D square+2D+5 annihilates

-2 times e to the -x times sine of 2x, right?

You can confirm it because, okay, think about that,

4:39

Can you consider this corresponding characteristic equation r squared + 2r

+ 5 = 0, which has two complex conjugate roots,

say -1 +- 2i, okay?

And that implies the general solution of this

differential equation y = C1 e to the -x cosine of

2x + C2 e to the -x and the sine of 2x, right, okay?

And that means what, okay?

D squared + 2D + 5 annihilates both e to the -x cosine 2x,

and e to the -x and sine 2x, right?

So okay, we can claim that this second of the polynomial in D, okay?

D squared +2D + 5 is an annihilator of the given function,

-2e to the -x and sine 2x, right, okay?

As a last example, okay,

consider the 3 times e to the -x sine

2x- 2x times e to the -x sine 2x + x

square e to the -x cosine 2x, okay.

In the example b, we have already seen that,

okay, D squared + 2D + 5, okay,

annihilates both e to the -x cosine 2x and

e to the -x sine 2x, right?

Okay, so, okay, this operator, this D square + 2D

+ 5 annihilates this first part, e to the -x, sine 2x, right?

Then what's the annihilator of x times e to the -x sine 2x, right?

Okay, we can see that, okay, D square + 2D + 5

square annihilates x times e to the -x and sine 2x, right?

By the same token, if we multiply one

more x to get x squared times e to the -x and

cosine 2x, then last term is annihilated

by D squared + 2D + 5, cubed, right?

So in fact, D squared + 2D + 5, cubed,

annihilates all the other three parts, so

it also annihilates the linear combination, okay?

So we found that finally D squared + 2D + 5, cubed,

is an annihilator of all these expression down here, okay.

7:41

Generalizing all those examples, we can see rather easily then,

for any positive integer n, and for any real numbers alpha and

beta, where beta is non-zero, okay.

D to the n annihilates x to the k,

where k is moving from 0 to n-1.

So that means D to the n also now annihilates,

any polynomial of degree and most n-1, okay?

8:57

Two numbers, right?

So product of these two expression and to the n, okay, it annihilates, okay?

Or expanding it out, okay, you will get D square-

2 alpha D + alpha squared + beta squared and

to the n, it annihilates, okay.

Also such expressions like x to the k, e to the alpha x,

cosine beta of x, or x to the k, e to the alpha x,

sine beta over x, where k is moving from 0 to n-1, okay?

Also at the same operator annihilates any

possible linear combination of all these expressions.